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Pauli matrices problem

  1. Jun 24, 2011 #1

    Given the two relations below, is it true and if yes, can anyone help me show that the solution to this must be the Pauli matrices? The alphas are matrices here.

    [itex]\alpha_{i}\alpha_{j}+\alpha_{j}\alpha_{i} = 2\delta_{ij}*1[/itex]. 1 is the identity matrix

    [itex]\alpha_{i}^{2} = 1[/itex]

    Thank you
  2. jcsd
  3. Jun 24, 2011 #2


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    If {[itex]\alpha_i[/itex]} is a solution, then {[itex]-\alpha_i[/itex]} is also a solution, so you matrices are not unique.

    You can explicitly calculate the constraints on the matrix entries if you start with [itex]\alpha_i = (a_i b_i c_i d_i)[/itex].
  4. Jun 24, 2011 #3
    I realize my post was not clear. My question is not how to get the Pauli matrices from the relations, that I know how to do. But how can I prove (if it is true) that any solution to these equations is a Pauli matrix upto a unitary transformation?
  5. Jun 24, 2011 #4
    It's not quite true but it's quite close to being true. (Notice that your second condition is redundant since it's contained in the first).

    If the matrices are 2x2, then they must be the Pauli matrices up to a similarity transformation. That is to say, there exists an invertible matrix S such that [itex]S\alpha_i S^{-1}[/itex] are the Pauli matrices.

    To show it:
    (1) You can prove that the matrices must be traceless. (Try it! There's a clever but very simple trick you can use.)
    (2) The matrices must have eigenvalues plus or minus 1.
    (3) Since they are 2x2, the eigenvalues are 1 and -1, once each.
    (4) Pick one of the matrices, say [itex]\alpha_3[/itex], and by similarity transformation make it diagonal so it is the third Pauli matrix.
    (5) Pick another of the matrices, say [itex]\alpha_2[/itex], and constrain its elements by the algebra relations. Use the residual symmetry transformation (i.e. the transformation that leaves [itex]\alpha_3[/itex] invariant) to make this equal to the first Pauli matrix.
    (6) Pick a third matrix. You'll find there is only one thing it can be. So that's as far as you can go.

    It's a fun (and relevant) exercise to do this for matrices of arbitrary size. There's a really nice sort of inductive argument.
  6. Jun 24, 2011 #5
    Ok I've figured out (1). Woohoo!

    Multiplying the given relation by [itex]\alpha_{j}[/itex] and taking the trace on both sides we get

    [itex] Tr(\alpha_{i}\alpha_{j}\alpha_{j}+\alpha_{j}\alpha_{i}\alpha_{j}) = Tr(0*\alpha_{j})=0[/itex]

    So [itex] Tr(\alpha_{i}) = -Tr(\alpha_{j}\alpha_{i}\alpha_{j})[/itex] since [itex]\alpha_{j}^{2} = 1[/itex]

    Since [itex]Tr(ABC) = Tr(CBA)[/itex] [itex]Tr(\alpha_{i}) = -Tr(\alpha_{i}) [/itex] which means that the trace must be 0.

    Now, I'm stuck at (5).

    Using the fact that my matrices are Hermitian and the third Pauli Matrix, I know that the matrices I have are of the form

    [tex]A=\begin{pmatrix} 0 & a \\ a^{*} & 0 \end{pmatrix}[/tex]

    Since the square of the matrix must be the identity, aa*=1 so a=1 or a=i.

    I don't understand what you mean by using the residual symmetry transformation to make this equal to the first Pauli matrix.

    Thank you
    Last edited: Jun 24, 2011
  7. Jun 24, 2011 #6
    Nicely done, that's the trick!

    When I say 'residual symmetry' I mean that when we diagonalised in part (4) the choice of S wasn't actually unique, and we still have some freedom left over. Specifically, we can choose another S as long as it commutes with the third Pauli matrix. If you work out the form that this must take, and what it does to your matrix A, you'll find that it's a rescaling of a. So we can use this to rescale a to be 1, so we get the first Pauli matrix.

    Minor point: you've chosen to constrain your matrices to be hermitian, so the symmetry transformations S are unitary. I didn't include that constraint, so I additionally show that we may take the matrices to be hermitian. With your assumption, the rescaling must in fact be just a phase, which figures because hermiticity constrains |a|=1.
  8. Jun 24, 2011 #7
    Sorry but I'm still not quite getting the point you're trying to make.

    First, is S one of the Pauli matrices or is it the unitary/similarity transformation matrix? What exactly do you mean by rescaling and how exactly should I show that the form of my [itex]\alpha_{2}[/itex] is always a rescaling of A?

    Thank you.
  9. Jun 25, 2011 #8
    Sorry I'm being a bit terse, I'll try to explain what I'm doing more fully.

    We start with a set of two by two matrices satisfying the algebra (I'm not going to require them to be hermitian, but you could). What we want to prove is that there is a similarity transformation S (again, not required to be unitary) such that [itex]S\alpha_i S^{-1}[/itex] are in the standard form of the Pauli matrices. The vital point is that the algebraic relations are unchanged under similarity transformations.

    The way we do this is by slowly building up the Pauli matrices one at a time, and constructing S as a product.

    At the first stage, we pick a similarity transform [itex]S_1[/itex] such that [itex]S_1\alpha_3 S_1^{-1}[/itex] is the diagonal Pauli matrix [itex]\sigma_3[/itex]. But there is nonuniqueness in this choice, in that we can pick [itex]S_2[/itex] such that [itex]S_2\sigma_3 S_2^{-1}=\sigma_3[/itex].

    Now at the second stage, we want to put a second matrix into standard form. But any further similarity transformation we do must not disturb the matrix we have already constructed, so it must be of the form [itex]S_2[/itex] (commute with [itex]\sigma_3[/itex]). This is what I meant when I said 'residual transformation'.

    Specifically, we find that [itex]S_2[/itex] must be diagonal. Since the overall scale of the similarity transformation is irrelevent, we can pick it to have the form [itex]\mathrm{diag}(\lambda,1)[/itex] for some nonzero [itex]\lambda[/itex]. The algebra, as you have said, constrains [itex]\alpha_1[/itex] to be of the form
    0 & a \\
    a^{-1} & 0
    and by calculation applying similarity transformation [itex]S_2[/itex] to this just rescales [itex]a\to\lambda a[/itex]. So pick [itex]\lambda=a^{-1}[/itex] and you get the Pauli matrix [itex]\sigma_1[/itex].

    Now the only similarity transformations that fix both the matrices we have constructed are proportional to the identity, so we have no freedom left. But the algebra constrains the third matrix to be [itex]\sigma_2[/itex] so we've proved what we set out to. The transformation [itex]S=S_2S_1[/itex].
  10. Jun 25, 2011 #9
    Thank you for replying henry.

    Let me see if I've understood what you're saying correctly. You are trying to find a transformation [itex]S[/itex] (which you later find to be [itex]S=S_{2}S_{1}[/itex]) which can be applied to any of the three solutions so that [itex]S\alpha_{i}S^{-1}[/itex] is one of the three Pauli matrices. Is this the objective of constructing S?

    I can more or less follow the algebra but I'm a bit confused about how you are attacking the problem. Thank you very much for explaining it in such detail for me.
  11. Jun 25, 2011 #10
    Yes that sounds pretty much right. It might help a bit if I express it in a slightly different way:

    We start with some algebraic relations, in this case the anticommutation relations [itex]\alpha_i\alpha_j+\alpha_j\alpha_i=2\delta_{ij}[/itex]. We would like to represent these relations as linear operators on a vector space, for example to be operators in a quantum system. In this case, we are after 2-dimensional representations, which means the vector space must be 2-dimensional.

    The statement we are trying to prove is then that for any such representation, we can pick a basis of the vector space such that the operators are the Pauli matrices.

    The way we've proved it is to start with an arbitrary basis, and then construct the transformation matrix S to take us to the basis we're after.
  12. Jun 25, 2011 #11
    Right, so since I'm on the right track, what I'm not clear about is this:

    We took [itex]\alpha_{3}[/itex] and got the third Pauli matrix by [itex]S\alpha_{3}S^{-1}[/itex]. But now, when we take the second matrix, [itex]\alpha_{2}[/itex], it is in the form [itex]A[/itex], due to the algebraic constrains. Isn't it necessary to show first that [itex]S_{1}\alpha_{2}S_{1}^{-1}=\alpha_{2}[/itex]? Only after this, we can use [itex]S_{2}=diag(\lambda, 1)[/itex] on it to scale it. In other words, we ensured that [itex]S_{2}[/itex] does not disturb [itex]\sigma_{3}[/itex] but we also need to show that [itex]S_{1}[/itex] does not disturb [itex]\sigma_{2}[/itex] right?

    Thank you

    EDIT: Oops, I got it now! Sorry ignore the bit above, I realize we don't have to worry about it. Thank you loads henry_m! I finally understand.
    Last edited: Jun 25, 2011
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